Shota is a dangerous fellow who likes to go rock climbing in active volcanoes. One time, when he was $30$ meters below the edge of a volcano, he heard some rumbling, so he decided to climb up out of there as quickly as he could. He climbed up at a constant rate. After $4.5$ seconds, he was $7.5$ meters below the edge of the volcano. How fast did Shota climb?
Solution: Let's say that Shota climbed at a rate of $V$ meters per second. Then, he climbed $V\cdot T$ meters in $T$ seconds. In addition, we know that when he began to climb, his distance from the edge of the volcano was $30$ meters. The remaining distance Shota had to climb is found by taking the original distance and subtracting from it the distance Shota had already climbed. We can express this with the equation $R=30-V\cdot T$, where: $R$ represents Shota's remaining distance to climb (in meters) $V$ represents Shota's climbing speed (in meters per second) $T$ represents the time (in seconds) We know that after $4.5$ seconds $(T={4.5})$, Shota was $7.5$ meters below the edge of the volcano $(R={7.5})$. Let's plug these values into the equation to find the value of $V$. $ \begin{aligned}{7.5}&=30-V\cdot{4.5}\\ 4.5V&=22.5\\ V&=5\end{aligned}$ Therefore, Shota climbed up at a rate of $5$ meters per second. To find how long it took Shota to reach the edge of the volcano, we can plug $R=0$ into the equation and solve for $T$. $ \begin{aligned}0&=30-5T\\ 5T&=30\\ T&=6\end{aligned}$ Shota climbed up at a rate of $5$ meters per second. It took Shota $6$ seconds to reach the edge of the volcano.